Download MODULE FOR INTERNATIONAL STANDARD CLASS

MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
DATE – TIME
TEACHER
SUB. THEME
: Chemistry
: …………………………………..
: Undang Suryana, HS.
: Hydrocarbons ( Isomerism & Reactions Of
Hidrocarbons )
CLASS/SEMESTER,YEAR: X/2, 2005 – 2006
I.
General Objectives
The students can understand organic compounds and molecul macro, determine
product of reaction and synthesise molecul macro as its using.
II.
Spesific Objectives
After completing the lessons, students are expected to be able to :
- explain isomerism concept & apply of classity of hydrocarbon compounds.
- Write simple reaction of alkanes, alkenes, alkynes.
III.
Materials
- ISOMERISM Isomerism is a hydrocarbon compound has same molecule formula and different
structural.
Example :
C4H8 → a) CH3 – CH2 – CH2 – CH3 = butane
b) CH3 – CH – CH3 = Isobutane
∣
CH3
- REACTIONS OF HYDROCARBONDS –
1. Oxidation
CH4 (8) + O2 ( 8) → CO2 = 2H2O
2. Substitution Reactions
CH4 + Cl2 → CH3Cl + HCl
3. Addition Reactions
CH3 CH = CH2 + Br2 → CH3 CH – CH2
∣
∣
Br
Br
4. Elimination Reactions
CH3 – CH – CH2 + Zn → CH3 – CH = CH2 + Zn Cl2
∣
∣
propen
Cl
Cl
IV.
Method / Technique
Demonstration
: Use molymood
V.
Competence Test
1) Make isomerism from C6H14, C6H12, C6H10
2) Write the following equestion
a. C3H8 + O2 →
b. Propene + H2 →
c. 2 – pentene + HCl →
d. 2 – meter – 2 – pentene + HI →
e. 2 cloro butane + KOH →
Refferences
Ebbing, General Chemistry, USA : Houghton Mifflin Company, 1987 Purba, Michael,
Kimia untuk SMA 1b, Jakarta : Erlangga 2004.
I.
Give the IUPAC name for each of the following hydrocarbon
1) CH3 – CH – CH – CH – CH3
∣ ∣ ∣
CH3 CH3 CH3
2) CH3 – CH – CH2 – CH2
∣
∣
C 2 H5
C2H5
3) CH3 – CH – CH2 = CH2 – CH – CH3
∣
∣
CH3H7
CH3
C2H5
∣
4) CH3 – C ≡ C - C – CH2 – CH – CH3
∣
∣
CH3
CH3
5) CH2 = CH – CH2 – C = CH – CH2
∣
∣
CH3
CH3
II.
Give the condesed structural formula for each of the following compounds
6) 2, 36 – trimetil oktana
7) 2,4 – dimetil – 1 – pentena
8) 4 – metil – 2 – pentuna
9) 3,3 – dietil – 4 – propil – 1,5 – heksadiena
10) Isobutana
III.
Make isomerism from C6H10
MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
DATE – TIME
TEACHER
SUB. THEME
: Chemistry
: …………………………………
: Undang Suryana, HS.
: Auto Redox
I.
General Objectives
Students explain the development of redox concept and relation with name system
of compound.
II.
Spesific Objectives
After completing the lessons, students are expected to be able to :
- Give the name IUPAC of compound
- Explain the auto redox reaction
III.
Materials
- Otoredox reaction was called : disproporsionasi.
- Otoredox reaction : redoks eaction which the same substance as redactor and
As oxidator.
Example :
1. Cl2 (g) + 2 NaOH (g) → NaCl (ag) + NaClO (ag) + H2O
reduksi
-1
oksidasi
+1
Past of Cl2 gas (biloks = 0) realize reduction to NaCl (biloks Cl = -1)
and another part realize oxidation to NaCl O (BILOKS Cl = + 1)
Cl2 = reduktor
Cl2 = oksidator
+4
2. H2S
-2
+ SO2
reduction o
→
oksidation
3S + 2 H2O
o
SO2 = oksidatur
H2S = reductor
-
The name system of Ionic Compound by IUPAC
Oxidation number can used for indicate “muatan” ionic metal in the ionic
comp.
Exp :
 CO2S → 2 CO + S2Biloks Cu = +1
Name : Tembaga (I) sulfide

Fe2 (SO4)3 → 2 Fe 3+ + 3 SO4 2biloks Fe = + 3
name : besi (III) sulfat

Pb (NO3)2 → Pb 2+ + 2NO3Biloks Pb = +2
Name : Timbal (II) nitrat
IV.
Competence Test
Answer the question !
1. Define auto redox or not, these reaction bellow ?
a. 3 NaClO → 2NaCl + NaClO3
b. 5 KI + KIO3 + 3 H2SO4 → 3 K2SO4 + 3 I2 + 3 H2O
2. Give the name these compound bellow :
a. SnO
b. CO2O
c. Al2 (SO4)3
d. HgCl2
e. MgO
V.
Reference
- Michael Purba, Kimia SMA 1B, Michael Purba.
VI.
Student’s difficulties and suggestions
MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
DATE – TIME
TEACHER
SUB. THEME
: Chemistry
: ………………………………………
: Undang Suryana, HS.
: Name System Of The Complex Ionic
I.
General Objectives
Students understand about property of the transisi element through obaservation
and concept applied.
II.
Spesific Objectives
After completing the lessons, students are expected to be able to :
- Give name of compleks ionic.
- Give name of coordinasi compound
III.
Materials
The name system of complex ionic, differentiate :
1. Complex Positive
Exp : coordinate number + ligan + name atomic centre + oxidation number
Zn (NH3)4 2 + = tentra aming seng (II)
Ag (NH3)2 + = di amin perak (I)
Cu (H2O)4 2+ = tetra aguo tembaga (II)
2.
Complex Negative
Exp : Koordinate number + ligan + name atomic centre + “at” + oxidation
number
Ag (CN)2 - = disiano Argentant (I)
Fe (CN)6 4 - = Heksa siano ferat (II)
CoF6 3= Heksa Floura Cobaltat (III)
IV.
Competence Test
Answer the question
1. Write the name of :
a). Zn (NH3) Cl +
b) Cr (H2O)4 Cl2 +
c) Cu (NH3)4 2+
2. Write chemistry formula of :
a). Kalium Heksasiano marganat (III)
b). Natrium tetrasianozinkat (II)
C). Tetra amin dikloeakobalt III nitrat
V.
Reference
- Michael Purba, Kimia 2000 3B, Erlangga
VI.
Student’s Difficulties and Suggestions
MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
DATE – TIME
TEACHEAR
SUB. THEME
CLASS / SEMESTER, YEAR
: Chemistry
: …………………………………
: Undang Suryana, HS.
: Petroleum & Petrochemical
: X / 2, 2005 - 2006
I.
General Objectives
Describe process of forming & technique of saparation of petroleum fraction with
Its use.
II.
Spesific Objectives
- Explain process of forming petroleum and natural gas.
- Explain main components of composing petroleum.
- Interpret of diagram of distilling stratified for explaining base and technique
of saparation in petroleum fraction.
- Differ gasoline quality base on octan number.
- Explain of using of petroleum residue in petrochemical industry.
- Analyze effect of fuel burning of environment.
III.
Materials
1. Petroleum composition
2. Petroleum refining
Petroleum, as we noted earlier, is a mixture of hydrocarbons, principally
alkanes and cycloalkanes. The object of petroleum refining is to obtain
various hydrocarbon products from the mixture. In part, this is accomplished
by fractional distillation of the petroleum, but the demand for certain products,
particularly gasoline, is greater than what can be supplied by distillation. For
the reson, petroleum refiners resort to various chemical processes is catalytic
cracking.
The antiknock characteristics of a gasoline are rated by the octane number
scale.
3. Process of petroleum.
4. Quality of gasoline.
IV.
Method / Technique
Demonstration : Use diagrams of petroleum.
IV.
Competence Tests
Answer the questions below !
1. How process of petroleum ?
2. What is meaning ?
a. Stratified distilation
b. Petroleum cracking
c. Octan number
3. How to improve petroleum quality ?
4. Mention additive substance in fuel ?
And mention profit and as unprofit ?
References
Purba, Michael, Kimia untuk SMA 1b, Jakarta : Erlangga 2004.
Ebbing, General Chemistry USA : Houghton Mifflin Company, 1987.
MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
: Chemistry
DATE – TIME
: ………………………………
TEACHER
: Undang Suryana HS.
SUB. THEME
: Reaction of Stoichiometry
CLASS/SEMESTER, YEAR: X (Science) / 2; 2005-2006
I.
General Objectives
The students can apply the law of Avogadro and Gay iussac with mol concept in
solving stoichiometry.
II.
Specific Objectives
After completing the lessons, students are expected to be able to :
 Determine limiting reagent in a reaction.
III.
Materials
Stoichiometry is the study of mole, mass, energy, and volume relationship
In chemical reactions. In stoichiometry, we usually look at the quantities of
reactans that combine together to produce various amount of products.
Example :
1) Calculate the theoretical maximum number of moles of NH3 that results when
0,55 mol H2 combines with excess N2 ? N2 + 3H2 → 2NH3
solutions
Ar C = 1 mol C X125 mol H2O x 12 g C = 15000 C.
1 mol H2O
Limiting Reagent Problems
Is reactant limit in reaction and the maximum yield of product.
Example:
1) What mass of SO3 is produced when 0,6 g SO2 is combined with 0,4 g 02 ?
2SO2 (g) + O2 (g) → 2SO3 (g)
Solution
2SO2(G) + O2(g) → 2SO3(g), Mr SO3 = 80
mol So2
= 0,68 SO2 = 0,00938 mol SO2 → limiting reagent
64g SO2
mol O2
= 0,4g SO2 = 0,0125 mol O2
32g O2
mass of SO3 = 2mol SO3 X 0,00938 mol SO2 x 80 gr SO3 = 0,75 gr SO2
2 mol SO2
IV.
Competence Test.
Answer the question below !
1) What mass of oxygen gas O2, is liberated when a 2,5 g sample of sodium
Nitrate is heated ? 2 NaNO3(S) → 2 Na NO2(S) + O2(g)
2) What volume of SO2, is liberated when 884 g S8is combined with excess O2 ?
S8(s) + 8 O2 (g) → 8 SO2(g)
3) Consider the equition : H2 + I2 → 2HI
Calculate the mass of HI that forms when 1,05 g H2 combine with 122 g I2 ?
V.
References
Drew H. Wholfe, introducing the college chemistry, US: Mc. Graw Hill, 1984.
Purba, Michael, Kimia untuk anak SMA kelas X Jilid 1 b, Jakarta : Erlangga 2002
VI.
Student’s Defficulties and Suggestions
MODULE FOR INTERNATIONAL STANDARD CLASS
SMK NEGERI 1 TAROGONG KALER
SUBJECT
: Chemistry
DATE-TIME
: …………………………………..
TEACHER
: Undang Suryana, HS.
SUB. THEME
: Compounds of Stoichiometry
CLASS/SEMESTER, YEAR: X (Science) / 2; 2005 – 2006
I.
General Objectives
The students can apply the law of Gay Lussac and Avogadro with mol concept in
Solving stoichiometry.
II.
Spesific Objectives
After completing the lessons, students are expected to be able to :
 Determine empirical formula from the composition and molecular formula
from empirical formula and mass percentages from the formula.
III.
Materials
1. An Empirical Formula or Simplest Formula from the Composition.
The empirical formula of a compound shows the ratios of number of atoms in
the compound.
Example :
A compound of nitrogen and axygen is analyzed, and a sample waging 1,87 g
is found to contain 0,483 grams N and 1,104 grams O. What is the empirical
formula of the compound ?
Solution :
Mol N = 0,483grams = 0,0345 mol N
14grams N
Mol O = 1,104grams = 0,06900 mol O
16gramsO
N
:
O
0,0345 :
0,069
So the empirical Formula : NO2
1
:
2
2. Molecular Formula from Empirical Formula.
The molecular formula of a compound is multiple of its empirical formula
n
=
molecularweight
empiricalformulaweight
Example :
We found the percentages compositions of butyric acid to be 54,2 % C, 9,2 %
H, and 36,6 % O. Determine the empirical formula. The molecular weight of
Butyric acid was determined by experiment to be 88 amu.What is the
molecular formula ?
Solutions :
C
:
H
54,2
:
12
4,51
:
1,97
:
2
:
(C2H4O) n = 88
(44)
n = 88
n = 2
:
O
9,2
1
9,2
4
4
:
36,6
16
:
2,29
:
1
:
1
So the molecular formula of butyric acid is
(C2H4O)2 or C4h8o2
3. Percentages of Element in Compound
% = x.Ar x 100 %
Mr
Example :
Calculate the percentage of C and N in CO (NH2)2 ?
Known: Ar H = 1, C = 12, O = 16
Solutions :
Mr CO (NH2)2 = 60
% C = 1(12) X 100 5 = 20 %
60
%N =
2(14) X 100 % = 46,67 %
60
IV.
Compeyence Tests
1. Benzoid acid is a white crystalline powder uae as a food preservative. The
Compound contains 68,8 % C, 5,0 % H, nad 26,2 % O by mass. What is its
Empirical formula ?
2. The percentage composition of etyl is 62 % C, 10,4 % H and 27,6 % O and
The molecular weight is 116 amu. Obtain the molecular formula.
V.
References
Drew H. Wolfe, IN TRODUCTION TO COLLEGE Chemistry, USA: pany, 1987.
Purba, Michael, KIMIA UNTUK sma Kelas I Jilid 1b. Jakrta: Erlangga, 2002.